How to find the solution to an equation - ACT Math

Start by plugging in x = 4 to solve for y: y = 3 * 4 + 5 = 17. Then 2 * 17 – 1 = 33

The best way to solve this problem is to turn the two statements into equations calling Sarah’s age S and Ron’s age R. So, S = 3(R – 2) and S = 14 + R. Now substitute the value for S in the second equation for the value of S in the first equation to get 14 + R = 3(R – 2) and solve for R. So R equals 10 so S equals 24 and the sum of 10 and 24 is 34.

Set each component of this equation equal to zero and solve for d and c.

(x-d)(x+c)=0

x-d=0

d=0

x+c=0

x=-c

Set up an equation to represent the total cost in cents: 24P + 76T = 652. In order to reduce the number of variables from 2 to 1, let the # tomatoes = 12 – # of potatoes. This makes the equation 24P + 76(12 – P) = 652.

Solving for P will give the answer.

The goal in this problem is to have only one variable. Variable “x” can designate Claire’s age.

Then Nick is x + 3, Kim is 2x, and Emily is 2x – 6; therefore x + x + 3 + 2x + 2x – 6 = 81

Solving for x gives Claire’s age, which can be used to find Nick’s age.

If we solve the equation for b, we add 2g to, and subtract 3h from, both sides, leaving 3h = 6g. Solving for h we find that h = 2g.

If we solve the first equation for 2x we find that 2x = 9 – y. If we solve the second equation for z we find z = –4 + y. Adding these two manipulated equations together we see (2x) + (y) = (9 – y)+(–4 + y).

The y’s cancel leaving us with an answer of 5.

We must find a common denominator and here they changed the first fraction by removing a negative from the numerator and denominator, leaving –11/(7 – x). We add the numerators and keep the same denominator to find the answer.

To solve for x:

bx + c = e – ax

bx + ax = e – c

x(b+a) = e-c

x = (e-c) / (b+a)

1. Substitute y in the equation for 4.
2. You now have 6 * 4 = 10z + 4
3. Simplify the equation: 24 = 10z + 4
4. Subtract 4 from both sides: 24 – 4 = 10z + 4 – 4
5. You now have 20 = 10z
6. Divde both sides by 10 to solve for z.
7. z = 2.

This goes up at a constant number between values, making it an arthmetic sequence. The first number is 2, with a difference of 3. Plugging this into the arithmetic equation you get An = 2 + 3 (n – 1). Plugging in 20 for n, you get a value of 59.

Looking at the sequence you can see that it doubles each term, making it a geometric sequence. Since it doubles r = 2 and the first term is 5. Plugging this into the geometric equation you get An = 5(2)n–1. Setting n = 6, you get 160 as the 6th term.

The zeroes of the equation are where f(x) = 0 (aka x-intercepts). Setting the equation equal to zero you get x2 = 9. Since a square makes a negative number positive, x can be equal to 3 or –3.

In order to solve for the x value you set both equations equal to each other (0.5x+3=3x-2). This gives you the x value for the point of intersection at x=2. Plugging x=2 into either equation gives you y=4.

First find 2%3 = (2 * 3 + 3 * 2)/(6 * 2 * 3) = 12/36 = 1/3, then 3%2 = (2 * 2 + 3 * 3)/(6 * 3 * 2) = 13/36 which is greater.

Starting with 12x + 3 = 2(5x + 5) + 1, we start by solving the parenthesis, giving us 12x + 3 = 10x + 11. We then subtract 10x from the right side and subtract three from the left, giving us 2x = 8; divide by 2 → x = 4.

The equation is xy + 5 = 69, making xy equal to 64. If we factor 64, we see that 1x64, 2x32, 4x16 and 8x8 all equal 16 when the two numbers are added together, so 24 is the only possibility that does not work.

5 more than 5 = 10

5 + x = 10

Subtract 5 from each side of the equation: x = 5 → 2_x_ = 10

Define the variables as x = the first number, x + 1, the second number, and x + 2 the thrid number.

The sum becomes x + x + 1 + x + 2 = 15 so 3x + 3 = 15. Subtract 3 from both sides of the equation to get 3x = 12 → 3x/3 = 12/3 → x = 4

The three numbers are 4, 5, and 6 and their product is 120.